力扣64. 最小路径和

class Solution {
    public int minPathSum(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0].length == 0) return 0;
        int rows = grid.length;
        int columns = grid[0].length;
        int[][] dp = new int[rows][columns];
        dp[0][0] = grid[0][0];
        for(int i = 1; i < rows; i++){
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        }
        for(int j = 1; j < columns; j++){
            dp[0][j] = dp[0][j - 1] + grid[0][j];
        }
        for(int i = 1; i < rows; i++){
            for(int j = 1; j < columns; j++){
                dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
            }
        }
        return dp[rows - 1][columns - 1];
    }
}

归纳：
根据grid先把第一行和第一列的dp算出来（前一个点的dp+当前的grid，根据grid和dp动态规划出最小的dp